ベクトルの内積内積の定義と性質

学習環境

内積・外積・空間図形を通してベクトルを深く理解しよう (数学のかんどころ 1) (飯高茂(著, 編集)、中村滋(編集)、岡部恒治(編集)、桑田孝泰(編集)、共立出版)の第2章(ベクトルの内積)、2.1(内積の定義と性質)の問題2.1の解答を求めてみる。

\begin{array}{l} x + 2 y + 3 z = 0 \\ - 4 x + 6 z = 0 \\ x^{2} + y^{2} + z^{2} = 1 \end{array}

\begin{array}{l} z = \frac{2}{3} x \\ x + 2 y + 2 x = 0 \\ y = - \frac{3}{2} x \\ x^{2} + \frac{9}{4} x^{2} + \frac{4}{9} x^{2} = 1 \\ 36 x^{2} + 81 x^{2} + 16 x^{2} = 36 \\ x^{2} = \frac{36}{133} \\ x = \pm \frac{6}{\sqrt{133}} \\ y = \bar{+} \frac{9}{\sqrt{133}} \\ z = \pm \frac{4}{\sqrt{133}} \end{array}

（複号同順）

コード(Wolfram Language, Jupyter)

a = {1, 2, 3};
b = {-4, 0, 6};

Solve[{a . x == b . x == 0, x . x == 1}, x]

This system cannot be solved with the methods available to Solve.: This system cannot be solved with the methods available to Solve.

Solve[{a . x == 0, b . x == 0, x . x == 1}, x]

This system cannot be solved with the methods available to Solve.: This system cannot be solved with the methods available to Solve.

v = {x, y, z};

Solve[{a . v == b . v == 0, v . v == 1}, v]